Question: Multiply the following complex numbers: $({5+2i}) \cdot ({-4+3i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({5+2i}) \cdot ({-4+3i}) = $ $ ({5} \cdot {-4}) + ({5} \cdot {3}i) + ({2}i \cdot {-4}) + ({2}i \cdot {3}i) $ Then simplify the terms: $ (-20) + (15i) + (-8i) + (6 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -20 + (15 - 8)i + 6i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -20 + (15 - 8)i - 6 $ The result is simplified: $ (-20 - 6) + (7i) = -26+7i $